The inverse power method for the p(x)-Laplacian problem
--Manuscript
Draft--Manuscript Number: JOMP-D-14-00262R1
Full Title: The inverse power method for the p(x)-Laplacian problem
Article Type: Original Research
Keywords: p(x)-Laplacian; eigenpairs; inverse power method
Corresponding Author: Marco Caliari, Ph.D.
University of Verona Verona, ITALY Corresponding Author Secondary
Information:
Corresponding Author's Institution: University of Verona Corresponding Author's Secondary
Institution:
First Author: Marco Caliari, Ph.D.
First Author Secondary Information:
Order of Authors: Marco Caliari, Ph.D.
Simone Zuccher, Ph.D. Order of Authors Secondary Information:
Abstract: We present an inverse power method for the computation of the first homogeneous
eigenpair of the p(x)-Laplacian problem. The operators are discretized by the finite element method. The inner minimization problems are solved by a globally convergent inexact Newton method. Numerical comparisons are made, in one- and
two-dimensional domains, with other results present in literature for the constant case p(x) ≡ p and with other minimization techniques (namely, the nonlinear conjugate gradient) for the p(x) variable case.
Link to official published version: https://www.sciencedirect.com/science/article/pii/S0377042716302989?via%3Dihub
Abstract We present an inverse power method for the computation of the first homogeneous eigenpair of the p(x)-Laplacian problem. The operators are discretized by the finite element method. The inner minimization problems are solved by a globally convergent inexact New-ton method. Numerical comparisons are made, in one- and two-dimensional domains, with
other results present in literature for the constant case p(x)≡ p and with other minimization
techniques (namely, the nonlinear conjugate gradient) for the p(x) variable case.
Keywords p(x)-Laplacian· eigenpairs · inverse power method
Click here to download Manuscript: pLaplace.tex Click here to view linked References
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2(will be inserted by the editor) M. Caliari, S. Zuccher
The inverse power method for the p(x)-Laplacian problem
M. Caliari · S. Zuccher
November 20, 2014
1 Introduction
In this paper we consider the generalization of the p-Laplacian eigenvalue problem with homogeneous Dirichlet boundary conditions
−∆pu(x) = ηp|u(x)|p−2u(x) (1)
where ∆ u(x) = div(|∇u(x)|p−2∇u(x)), Ω is a bounded domain in Rd, u
∈ W01,p(Ω ) and
p> 1, to the case with variable exponent p(x).
The p-Laplacian operator appears in several mathematical models that describe nonlin-ear problems in physics and mechanics [7, 9]. Examples include fluid dynamics [11], mod-eling of non-Newtonian fluids and glaciology [4, 10, 12, 20, 27, 28], turbulent flows [15], climatology [16], nonlinear diffusion (where the equation is called the N-diffusion equa-tion, see Ref. [29] for the original article and Ref. [21] for some current developments), flows through porous media [30], power law materials [5] and torsional creep [24].
The p-Laplacian operator where p is replaced by p(x), 1 < p−≤ p(x) ≤ p+< +∞, is
deeply related to generalized Lebesgue and Sobolev spaces, which have been vigorously
studied and whose theory has been ripe for applications to PDEs.The common
assump-tion in literature is that the exponent p(x) is a measurable funcassump-tion and 1/p(x) is globally log-H¨older continuous [17].There have been many contributions to nonlinear elliptic prob-lems associated with the p(x)-Laplacian from various view points (see Ref. [1, 2, 14] and Ref. [22] for a survey), whereas there are much less contributions to parabolic problems [3] and eigenvalue problems [18].
If p is constant, it is well known that the smallest eigenvalue satisfies
ηp= inf u∈W01,p(Ω ) u6=0 R Ω|∇u| p dx R Ω|u| p dx . (2) M. Caliari· S. Zuccher
Department of Computer Science, University of Verona, Strada Le Grazie 15, 37134, Verona, Italy M. Caliari E-mail: marco.caliari@univr.it S. Zuccher E-mail: simone.zuccher@univr.it 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51
The trivial generalization of quotient (2) to the case p= p(x) yields a problem in which
the homogeneity is lost, that is if u is a minimizer then ωu, ω6= 0, is not. In Ref. [19], the
homogeneity is restored considering
λ= inf u∈W1,p(x) 0 (Ω ) u6=0 k∇ukp(x) kukp(x) ,
where the normk·kp(x)is the so called Luxemburg norm
kukp(x)= infγ>0 ( γ : Z Ω u(x) γ p(x) dx p(x)≤ 1 ) . (3)
The use of p(x)−1dx (rather than the classical dx) just simplifies the equations a little (see Ref. [19]).The corresponding Euler–Lagrange equation is
−div ∇u(x) k∇ukp(x) p(x)−2 ∇u(x) k∇ukp(x) = λ T u(x) kukp(x) p(x)−2 u(x) kukp(x) , (4) where T= R Ω ∇u(x) k∇ukp(x) dx R Ω u(x) kukp(x) dx .
We notice that in the constant case p(x)≡ p, since pkukp
p=kuk p Lp,
(λ , u) eigenpair for (4)⇔ (λp, u/√p p) eigenpair for (1) (5)
withkukp=
u/√pp
Lp= 1.
Although some different numerical methods for the eigenvalue problem with constant p are available, see Ref. [7, 9, 13], to the knowledge of the authors the first numerical method, based on the nonlinear conjugate gradient method, for computing
λ2= Λ = inf u∈W01,p(x)(Ω ) u6=0 k∇uk2 p(x) kuk2p(x) (6)
and the corresponding minimizer was proposed by the first author in Ref. [6], where only the case p(x)≥ 2 was considered. Here we use a preconditioned quadratic model for the minimization problem, either with the exact Hessian, if p(x)≥ 2, or with a modified Hessian, otherwise. The new method turns out to be more general, more robust and faster (see the numerical experiments in Table 3, Section 3) than the previous one.
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2 The inverse power method for a nonlinear eigenproblem
Given R : Rn→ R
+and S : Rn→ R+convex, differentiable, even and positively
2-homoge-neous functionals, with the further assumption that S(u) = 0 only if u = 0, then the critical
points u∗of F(u) = R(u)/S(u) satisfy
∇uF(u∗) = ∇uR(u∗)−
R(u∗)
S(u∗)∇uS(u∗) = 0.
If we define r(u) = ∇uR(u), s(u) = ∇uS(u) and Λ∗=R(u
∗)
S(u∗), then the critical point u∗satisfies
the nonlinear eigenproblem
r(u∗)−Λ∗s(u∗) = 0. (7)
If R(u) =hu,Aui, where A: Rn → Rn is linear and S(u) =hu,ui, h·,·i being the scalar
product in Rn× Rn, the standard linear eigenproblem is retrieved. The generalization of
the inverse power method for the computation of the smallest eigenvalues is the following algorithm:
1. take ˜u0random and compute u0= u˜
0 S( ˜u0)1/2 and Λ 0= F(u0) = R(u0) 2. repeat – ˜uk= arg min u {R(u) − hu,s(u k−1)i} – uk= u˜ k S( ˜uk)1/2 – Λk= F(uk) = R(uk) until convergence
It is proven [23] that the sequence{(uk,Λk)}
kproduced by the algorithm satisfies F(uk) <
F(uk−1) and converges to an eigenpair (u∗,Λ∗) for problem (7), with u∗a critical point of
F.
Let now u(x) be a linear combination of basis functions u(x) =
n
∑
j=1
ujφj(x),
with u= (u1, . . . , un)T the vector of coefficients, and let us apply the algorithm described
above to the case
R(u) = n
∑
j=1 uj∇φj(·) 2 p(x) , S(u) = n∑
j=1 ujφj(·) 2 p(x)In our numerical experiments, the set of functions{φj(x)}nj=1will be a proper set of finite
element basis functions.With abuse of notation, we will write for short R(u) =k∇uk2p(x)
and S(u) =kuk2p(x).
We remind that the algorithm does not guarantee the convergence to the smallest eigen-value. On the other hand, it is possible to apply it to different initial guesses and then take the smallest eigenvalue.
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2.1 Initial values and exit criteria
Although ˜u0can be chosen randomly, it is clear that the more it is close to u∗, the faster the
algorithm converges. In order to speed up the convergence, we use a continuation technique
over p(x), starting from p(x)≡ 2 for which the solution is known at least for common
domains Ω . By employing a convex combination
pm(x) = 2· (1 − θm) + p(x)· θm, θm=
r m
M, m= 0, 1, . . . , M, (8)
m-th problem is solved by providing as initial condition the solution of(m− 1)-th problem.
For the minimization problem (called inner problem) ˜
uk= arg min
u {R(u) − hu,s(u
k−1)i}, given uk−1
we use an iterative method as well. Since the inverse power method, at convergence, ensures
r(uk−1)−Λk−1s(uk−1)≈ 0 ⇒ r uk−1
Λk−1
− s(uk−1)≈ 0,
and (at each iteration)
r( ˜uk)
− s(uk−1) = 0,
a good choice for the initial value of ˜uk is uk−1/Λk−1. By the proof of Lemma 3.1 in
Ref. [23], we know that the descent in F is guaranteed not only for the exact solution of
the inner problem but also for any vector ¯ukwhich satisfies the condition
R( ¯uk) − h ¯uk, s(uk−1)i < R uk−1 F(uk−1) − uk−1 F(uk−1), s(u k−1) .
Since F(uk−1) = Λk−1, uk−1/F(uk−1) is precisely the above suggested choice for the initial
value of ˜uk. Therefore, in principle, starting from this value, the use of a descent method
for the inner problem would require a single step in order to guarantee the descent in F. More realistically, this means that far from convergence (for instance during the continuation technique) it makes no sense to solve the inner problem too accurately. For this reason, an exit criterion could encompass both a relatively large tolerance and a relatively small maximum number of iterations.
A classical exit criterion for the inverse power method is Λ k −Λk−1 ≤tol· Λ k (9)
for a given tolerance tol. As soon as it is satisfied, we assume that Λk≈ Λ∗and thus
r(uk)
−Λks(uk)
≈ 0. (10)
We call the left hand side residual; clearly, its infinity norm could be used as a different exit criterion. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51
2.2 Approximation of the Luxemburg norm
If u(x) is notzero almost everywhere, then its Luxemburg norm γ is implicitly defined by
F(u, γ) = Z Ω u(x) γ p(x) dx p(x)− 1 = 0. (11)
Hereafter we remove the integration domain Ω from the notation. The derivative with re-spect to γ of F(u, γ) is ∂γF(u, γ) =− 1 γ Z u(x) γ p(x) dx (12)
and thus Newton’s method can be easily applied. For a given u(x), F(u, γ) is a monotonically
decreasing convexC2function in γ. This guarantees the convergence of Newton’s method
whenever the initial guess γ0> 0 is chosen such that F(u, γ0)≥ 0. For the initial guess γ0,
we use the following estimates
γ0= Z Ω |u(x)|p(x) p(x) dx !pmax1 if Z Ω |u(x)|p(x) p(x) dx≥ 1 Z Ω |u(x)|p(x) p(x) dx ! 1 pmin if Z Ω |u(x)|p(x) p(x) dx< 1
where pmax= max p(x) and pmin= min p(x). In fact, if we consider the first case,
F(u, γ0) = Z Ω u(x) γ0 p(x) dx p(x)− 1 ≥ Z Ω |u(x)|p(x) γ0pmax dx p(x)− 1 = = 1 R Ω |u(x)|p(x) p(x) dx Z Ω|u(x)| p(x) dx p(x)− 1 = 0.
2.3 The inner problem
In the inverse power method, we need to solve the inner problem ˜
uk= arg min
u f(u) = arg minu {R(u) − hu,s(u
k−1)i}, given uk−1.
We first computehv,s(u)i = hv,∇uS(u)i for given vectors u and v. Since γ(u) = kukp(x)is
implicitly defined by F(u, γ(u)) = 0, the differentiation of implicit functions leads to
hv,∇uF(u, γ(u))i = hv,∂uF(u, γ(u))i + hv,∂γF(u, γ(u))∇uγ(u)i = 0,
from which
hv,∇ukukp(x)i = hv,∇uγ(u)i = −hv,∂
uF(u, γ(u))i
∂γF(u, γ(u))
. By recalling that, in general,
∂ui Z g(u(x))dx = Z g0(u(x))φi(x)dx, 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51
for v(x) = ∑n j=1vjφj(x) we have hv,∂ui Z g(u(x))dx i = Z g0(u(x))v(x)dx and thus hv,∂uF(u, γ)i = Z u(x) γ p(x)−2u(x) γ v(x) γ dx. By employing (12) we get hv,∇ukukp(x)i = R u(x) kukp(x) p(x)−2 u(x) kukp(x)v(x)dx R u(x) kukp(x) p(x) dx , from which
hv,s(u)i = hv,2kukp(x)∇ukukp(x)i = 2
R u(x) kukp(x) p(x)−2 u(x)v(x)dx R u(x) kukp(x) p(x) dx . (13)
Given uk−1of unitary Luxemburg norm, the problem reduces to find the minimizer of
f(u) = R(u)− hu,s(uk−1)i =
=k∇uk2p(x)− 2 R uk−1(x) p(x)−2uk−1(x)u(x)dx R |uk−1(x)|p(x)dx (14)
We use a line search method based on the sufficient decrease condition. If ucis the solution
at the current iteration, its update is
u+= uc+ δ d, (15)
where d is a descent direction, i.e.h∇uf(uc), di < 0, and δ is such that
f(uc+ δ d) < f (uc) + αδh∇uf(uc), di (16)
with α= 10−4. Condition (16) is usually called Armijo’s rule.
2.3.1 The quadratic model for the inner problem, p(x)≥ 2
We first consider the case p(x)≥ 2 and approximate f (u) in (14) with a quadratic model
f(u)≈ f (uc) +h∇uf(uc), u− uci +
1
2hu − uc, H(uc)(u− uc)i
where H(uc) is the exact Hessian of f at the current iteration uc. For p(x)≥ 2, R(u) is twice
continuously differentiable and convex, i.e. the Hessian of f is symmetric positive definite. Therefore
∇uf(u+)≈ ∇uf(uc) + H(uc)(u+− uc) = 0⇔ u+− uc=−H(uc)−1∇uf(uc)
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and hence the descent direction is d=−H(uc)−1∇uf(uc). If condition (16) is not satisfied
for a given δ , then it is repeatedly reduced (backtracking) until the condition is satisfied,
using a quite standard cubic polynomial model for p(δ ) = f (uc+ δ d) (see Ref. [26]).
Let us compute ∇uf(u) and H(u)v, for a given v∈ Rn. With the arguments used to
obtain (13), we immediately have
ri(u) = ∂uiR(u) = 2 k∇ukp(x) R ∇u(x) k∇ukp(x) p(x)−2 ∇u(x) k∇ukp(x)· ∇φi(x)dx R ∇u(x) k∇ukp(x) p(x) dx (17)
where here· denotes the scalar product in Rd×Rd. Since uk−1has unitary Luxemburg norm,
∂uif(u) = 2 KR ∇u K p−2 ∇u K · ∇φi R ∇u K p − 2 R uk−1 p−2 uk−1φi R |uk−1|p (18)
where we set K=k∇ukp(x) and omitted the dependency on x in the integrals in order to
reduce the notation. By introducing
Ni(u) = Z ∇u K p−2 ∇u· ∇φi, D(u) = Z ∇u K p
the i-th row of H(u)v is
(H(u)v)i= 2
D(u)hv,∇uNi(u)i − Ni(u)hv,∇uD(u)i
D(u)2 = = 2 R (p− 2) ∇u K p−4 ∇u K · ∇vK−∇uhv,∇uKi K2 ∇u· ∇φi+ ∇u K p−2 ∇v· ∇φi D(u) + − 2 Ni(u)Rp ∇u K p−2 ∇u K · ∇vK−∇uhv,∇uKi K2 D(u)2 = = 2 R (p− 2) ∇u K p−4 ∇u K · ∇v∇uK · ∇φi D(u) + − 2hv,∇u KiR (p− 2) ∇u K p−2 ∇u K · ∇φi D(u) + 2 R ∇u K p−2 ∇v· ∇φi D(u) + − 2 Ni(u) K R p ∇u K p−2 ∇u K · ∇v D(u)2 + 2 Ni(u) K hv,∇uKi R p ∇u K p D(u)2 .
By using again (13), the inner producthv,∇uKi can be recast as
hv,∇uKi = R ∇u K p−2 ∇u K · ∇v D(u) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51
and the i-th row of H(u)v becomes (H(u)v)i= 2 R (p− 2) ∇u K p−4 ∇u K · ∇v ∇u K · ∇φi D(u) + − 2 R ∇u K p−2 ∇u K · ∇v R (p− 2) ∇u K p−2 ∇u K · ∇φi D(u)2 + + 2 R ∇u K p−2 ∇v· ∇φi D(u) + − 2 R ∇u K p−2 ∇u K · ∇φi R p ∇u K p−2 ∇u K · ∇v D(u)2 + + 2 R ∇u K p−2 ∇u K · ∇φi R ∇u K p−2 ∇u K · ∇v R p ∇u K p D(u)3 . (19)
The reason for computing the action of H(u) to a vector is that H(u) is not a sparse matrix,
even if finite elements are used. Therefore, the descent direction−H(uc)−1∇uf(uc) is
com-puted through an iterative method such as the preconditioned conjugate gradient method,
which only requires the action of H(uc) to a vector v. A good choice for the preconditioner
in computing ˜ukcan be suggested by the modified gradient
∂uifc(u) = 2 Z ∇uc Kc p−2 ∇u· ∇φi Z ∇uc Kc p − 2 R uk−1 p−2 uk−1φi R |uk−1|p
and the corresponding modified Hessian Hc, whose entries are
2 Z ∇uc Kc p−2 ∇φj· ∇φi Z ∇uc Kc p . (20)
In fact, this preconditioner turns out to be sparse in case the basis functions{φj(x)}
cor-respond to a finite element method. It is possible to use a fixed preconditioner for each iteration of the inverse power method (and not a preconditioner for each iteration of the inner problem), starting from the modified gradient
∂uif k−1(u) = 2 Z ∇uk−1 Kk−1 p−2 ∇u· ∇φi Z ∇uk−1 Kk−1 p − 2 R uk−1 p−2 uk−1φi R |uk−1|p
and the corresponding modified Hessian Hk−1, whose entries are
2 Z ∇uk−1 Kk−1 p−2 ∇φj· ∇φi Z ∇uk−1 Kk−1 p . (21) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51
Once the modified Hessian is computed, it is factorized by Choleski’s method and the trian-gular factors are inverted in order to get the preconditioned conjugate gradient iterations.
2.3.2 The quadratic model for the inner problem, p(x)≥ p−> 1
If, for some values of x, p(x) is less than two and ∇u(x) is zero, than the first and the third term in the Hessian (19) are not defined, while the gradient (18) is still well defined, in the sense that
|∇u(x)|p(x)−2∇u(x)· ∇φi(x) = 0 if ∇u(x) = 0.
We can consider (see Ref. [9]) the modified Hessian arising from
Nεn i (u) = Z εn2+ ∇u K 2!p−22 ∇u· ∇φi, D(u) = Z ∇u K p
where εnis a small quantity, possibly depending on the number n of basis functions φj(x)
2 R (p− 2) εn2+ ∇u K 2p−42 ∇u K · ∇v ∇u K · ∇φi D(u) + − 2 R ∇u K p−2 ∇u K · ∇v R (p− 2) εn2+ ∇u K 2p−22 ∇u K · ∇φi D(u)2 + + 2 R εn2+ ∇u K 2p−22 ∇v· ∇φi D(u) + − 2 R εn2+ ∇u K 2p−22 ∇u K · ∇φi R p ∇u K p−2 ∇u K · ∇v D(u)2 + + 2 R εn2+ ∇u K 2p−22 ∇u K · ∇φi R ∇u K p−2 ∇u K · ∇v R p ∇u K p D(u)3
We notice that, in this way, the integrand function of the first term above side is equal to
zero where ∇u(x) = 0, and well defined otherwise, even without ε2
n. Therefore we replace εn2+ ∇u K 2 with (1− sign|∇u|2 ) + ∇u K 2
in that term. Indeed, such a modification is useful in the numerical implementation of f , its
gradient and its exact Hessian, even in the case p(x)≥ 2, for the evaluation of terms (see,
for instance, the second, the fourth and the fifth above) like
|z|p−2z, 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51
which can be numerically computed as
((1− sign|z|2) +|z|2)p−22 z.
On the other hand, if ∇u(x) = 0, the third term in the Hessian above is not defined without
εn2, but it is well defined if ∇u(x)6= 0. Therefore, we replace
εn2+ ∇u K 2
in the third term of the Hessian above and in the corresponding modified Hessian (see (20)) with εn2· (1 − sign|∇u| 2 ) + ∇u K 2 .
When p(x)→ 1+ for some values of x, the eigenvalue problem is numerically quite
difficult. This is due not only to the fact that the exact Hessian does not exist, but also to the high derivatives of the solution at the boundaries. In fact, as reported in Figure 1 for the case
of p constant, when p→ 1+the first eigenfunction on the ball converges to its characteristic
function (see Ref. [25]).
3 Numerical results 3.1 Details of the algorithm
For the computation of the Luxemburg norm, Newton’s method is stopped when the relative
difference of two successive solutions is smaller then 10·ε, where ε is the machine precision.
From our numerical experience, we found that a number of continuation steps in (8) chosen as M= max p(x) 2 if min p(x)≥ 2
d−3log(min p(x) − 1)e if min p(x) < 2
provides a satisfactory behavior. Moreover, the m-th problem in (8) is solved with tolerance
tolm= tol0· (1 − θm) + tol· θm
where tol0= 100· tol and tol is the tolerance in the exit criterion (9) for the inverse power
method. It is chosen proportional to hr+1, where r is the degree of the finite element
piece-wise polynomial basis. The same tolerance is used for inner problem (14). Four exit criteria are used. Since we use (modified) Newton’s method, if the infinity norm of the descent
di-rection δ d in (15) is smaller than tol· kuck∞already with δ= 1, we stop the iterations. Then
we consider the ratio between the relative change in f(u+) and the relative change in u+. If
∇uf(u+)◦ u+ f(u+) ∞ ≤ tol,
where◦ denotes Hadamard’s product, we stop the iterations. We stop the iterations if the
method stagnates, that is ifkδ dk∞≤ 0.1 · tolkuck∞, too. Finally, we fix the maximum
num-ber of iterations to 10. The value of εnfor the modified Hessian is 10−5.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51
3.2 One-dimensional case
We implemented and run the one-dimensional case in GNU Octave 3.8, using piecewise
linear basis functions φj(x) and approximating the arising integrals by midpoint quadrature
formulas.
The initial solution u0(x) corresponding to p(x)≡ p = 2 is the (piecewise linear
inter-polation of the) first eigenfunction of
−u00(x) = η2u(x), x0< x < x1
normalized in order to have u0
2= 1, that is u0(x) =√ 2 x1− x0 sin x− x0 x1− x0π , η2= η2= π x1− x0 2 . p λp ηp− λ p /ηp 1.1 1.28 8.6e-05 1.2 1.46 1.9e-05 1.3 1.61 4.8e-05 1.4 1.75 5.9e-05 1.5 1.88 6.5e-05 1.6 2.00 6.8e-05 1.7 2.12 7.1e-05 1.8 2.24 7.5e-05 1.9 2.35 7.8e-05 2.0 2.47 8.2e-05 3.0 3.54 1.6e-04 4.0 4.56 3.0e-04 5.0 5.58 5.1e-04 6.0 6.59 8.0e-04 7.0 7.59 1.2e-03 8.0 8.59 1.6e-03 9.0 9.59 2.1e-03 10.0 10.59 2.6e-03
Table 1 Values of λpand relative error with respect to ηpfor some values of constant p(x)≡ p, in the
interval(−1,1).
The first test aims at comparing our results with the analytical solution, which is known
for constant values of p(x)≡ p in the interval (−1,1) (see, for instance, Ref. [8])
ηp=
2√p p
− 1(π/p) 2 sin(π/p)
p
Table 1 reports the eigenvalue λp, computed with 101 basis functions, and the relative
differ-ence with the corresponding analytical value ηp(see (5)). Figure 1 shows the eigenfunction
u(x) for different values of constant p. In the limit p→ 1+it tends to an almost constant
function with steep gradients at the boundaries due to the homogeneous boundary
condi-tions, whereas in the limit p→ +∞ the eigenfunction resembles the absolute value of a
linear function (non differentiable at the origin).
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51
0 0.5 1 1.5 2 -1 -0.5 0 0.5 1 u (x ) x p = 1.1 p = 1.5 p = 2.0 p = 10
Fig. 1 First eigenfunctions for some values of constant p(x)≡ p.
0 10 20 30 40 50 60 -1 -0.5 0 0.5 1 p( x ) x 0 10 20 30 40 50 60 -1 -0.5 0 0.5 1 p( x ) x p(x)≡ 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 -1 -0.5 0 0.5 1 u (x ) x 0 0.2 0.4 0.6 0.8 1 1.2 1.4 -1 -0.5 0 0.5 1 u (x ) x
Fig. 2 p(x) = 28 + 26 cos(2πx) (left) and p(x) = 1.1 if x < 0 and p(x) = 10 if x≥ 0 (right) and the corre-sponding eigenfunctions (bottom).
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51
Figure 2 reports the first eigenfunction for two cases. In the first one (left), p(x) =
28+ 26 cos(2πx), in the second one (right) p(x) is discontinuous, p(x) = 1.1 if x < 0 and
p(x) = 10 otherwise.The second case, although does not satisfy the log-H¨older continuity
usually required to the exponent, was chosen as a limit case to check the robustness of the method.The values of λ for the two cases are about 0.94548 and 1.25999, respectively; the corresponding eigenfunctions are reported in the second row of plots. The values of residuals (10) in infinity norm are 0.0018 and 0.64, respectively. The second residual is
not so small. In fact, the first part of the numerical residual contains terms like|∇u(x)|p(x)−1
(see (17)), with ∇u(x) close to zero where p(x) is close to one. Those terms are approximated
by|∇u(x) + O(h)|p−1, where h the mesh size, which can be quite far from their exact values.
A possible confirmation of such an explanation is the fact that the maximum of the residual
is taken at ¯x= 0.04, where ∇u( ¯x)≈ 0 and p( ¯x) = 1.1.Similar values of the residuals were
observed for the smaller values of p in the constant case.
3.3 Two-dimensional cases
We implemented and run the algorithm with FreeFem++ 3.23, using piecewise quadratic elements for u(x, y) and piecewise constant elements for p(x, y), in order to better manage discontinuous functions p(x, y).
3.3.1 The rectangle
The initial solution u0(x, y) corresponding to p(x, y)≡ p = 2 is the first eigenfunction of
−∇2
u(x, y) = η2u(x, y), x0< x < x1, y0< y < y1
normalized to have u0 2= 1, that is u0(x, y) = s 8 (x1− x0)(y1− y0) sin x− x0 x1− x0 π sin y− y0 y1− y0 π η2= η2= π x1− x0 2 + π y1− y0 2 . p λp Ref. [13] Ref. [9] 1.5 10.0722 10.0722 10.0720 2.0 19.7393 19.7392 19.7392 2.5 35.9490 35.9493 35.9487 4.0 176.618 176.693 176.598 Table 2 Values of λpfor some values of constant p(x)≡ p.
For some constant values of p(x)≡ p, Ref. [13, Table 4, p-version] reports the values of
the eigenvalues λpwith six significant digits for the unit square and Ref. [9, Table 2] reports
the values for a wider range of p. In Table 2 we compare our results for p= 1.5, 2, 2.5 and
4.0 computed on a regular mesh with 26× 26 vertices and 1250 triangles. We find that our
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51
precNewton(10) precNLCG(10) λ residual CPU s. λ residual CPU s. p1(x) 4.0571 2.16e-4 38.46 4.0571 2.19e-4 84.42
p2(x) 4.1727 2.51e-4 49.32 4.1727 2.52e-4 77.14
Table 3 Comparison for the cases p1(x, y) = 5 + 3 sin(3πx) and p2(x, y) = 4.5 + 3 sin(3πx).
results are closer to those obtained by the p-version of FEM algorithm used in Ref. [13] than to those obtained in Ref. [9].
In order to test the code performance, we compared our new minimization method
(calledprecNewton(10)in Table 3) with the FreeFem++ functionNLCG, which implements
the nonlinear conjugate gradient minimization method (see Table 3). In a first example, taken
from Ref. [6, Section 2.2], p1(x, y) = 5 + 3 sin(3πx). Here, differently than in Ref. [6],we
implemented a preconditioner for theNLCGmethod based on modified Hessian (21) and set
the maximum number of iterations to 10. Therefore, the method is denotedprecNLCG(10)
in Table 3. In a second example we checked the case p2(x, y) = 4.5 + 3 sin(3πx), in which
p2(x, y) assumes values less than two. As shown in Table 3, our method outperforms the
preconditionedNLCGin terms of CPU time (with comparable residuals).
1 0.8 x 0.6 0.4 0.2 0 0 0 y 0.5 1 1.5 2 0.8 1 0.6 0.4 0.2
Fig. 3 Eigenfunction for p3(x, y). λ≈ 4.3491, residual 0.16.
As a final test, we considered a discontinuous case with
p3(x, y) =
(
1.1 if x2+ y2
≤ 1
10 otherwise
The corresponding eigenfunction u(x, y) is shown in Figure 3. Its shape resembles that of the one-dimensional case (see Figure 2, right).
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51
3.3.2 The disk
We computed the eigenvalues for the disk for some constant values of p(x)≡ p, namely
p= 1.5, 2, 2.5 and 4. The initial solution u0(x, y) corresponding to p(x, y)≡ p = 2 is the
normalized first eigenfunction of
−∇2
u(x, y) = η2u(x, y), x2+ y2< r2
that is u0(x, y) = u˜ 0(x, y) k ˜u0k 2 , u˜0(x, y) = J0 j0,1 p x2+ y2 r ! , η2= η2= j0,1 r 2
where J0is the Bessel function of the first kind of order zero and j0,1≈ 2.40482555769577
is its first zero, computed by the commands
f = @(z) besselj(0,z);
fzero(f,[2,3],optimset("TolX",eps))
in GNU Octave 3.8. In Table 4 we compare our results for the unit disk (r= 1), on a mesh
p λp Ref. [9] 1.5 4.0167 4.0053 2.0 5.7815 5.7616 2.5 7.7081 7.6815 4.0 14.6785 14.6369 Table 4 Values of λpfor some values of constant p(x)≡ p.
with 2032 vertices and 3912 triangles, with those provided in Ref. [9, Table 1] (there
re-ported with five or six significant digits). We notice that for p(x)≡ 2, the analytical solution
is j20,1≈ 5.7832 and that our value is more accurate.
3.3.3 The annulus
The initial solution u0(x, y) corresponding to p(x, y)≡ p = 2 for the annulus is the
normal-ized first eigenfunction of
−∇2
u(x, y) = η2u(x, y), a2< x2+ y2< b2
that is u0(x, y) = u˜ 0(x, y) k ˜u0k 2 , η2= η2= k0,1 a 2 where ˜ u0(x, y) = J0(k0,1)Y0 k0,1 p x2+ y2 a ! −Y0(k0,1)J0 k0,1 p x2+ y2 a !
If we choose a= 0.5 and b = 1, then k0,1≈ 3.12303091959569 is the first zero of
J0(z)Y0 zb a −Y0(z)J0 zb a
computed by the commands
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51
a=0.5;,b=1;
f = @(z) besselj(0,z).*bessely(0,z*b/a)-bessely(0,z).*besselj(0,z*b/a); fzero(f,[2,4],optimset("TolX",eps))
and Y0being the Bessel function of the second kind of order zero. The eigenvalues in Table 5
p λp exact
1.5 14.8932
2.0 38.9932 39.0133 2.5 95.0417
4.0 1154.08 Table 5 Values of λpfor some values of constant p(x)≡ p.
were obtained on a mesh with 1502 vertices and 2780 triangles. The exact value of η2for
the analytical known case p(x)≡ 2 is 4k2
0,1.
4 Conclusions
We described an effective method for the computation of the first eigenpair of the p(x)-Laplacian problem. The numerical results, in one- and two-dimensional domains, agree well
with those analytical or numerical available in literature for the constant case p(x)≡ p. The
main contribution of the present work is a fast and reliable method for the general case of
variable exponent p(x)≥ p−> 1, applied for the first time to some test problems in the
critical range p(x) < 2 somewhere in the domain.
Acknowledgements The authors cordially thank Dr. Stefan Rainer for the helpful discussions about numer-ical methods for unconstrained minimization.
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